Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
  ↳1 QTRS Reverse (⇔)
    ↳2 QTRS
  ↳3 QTRS Reverse (⇔)
    ↳4 QTRS
  ↳5 DependencyPairsProof (⇔)
    ↳6 QDP
    ↳7 DependencyGraphProof (⇔)
    ↳8 AND
      ↳9 QDP
        ↳10 UsableRulesProof (⇔)
          ↳11 QDP
          ↳12 UsableRulesReductionPairsProof (⇔)
          ↳13 QDP
          ↳14 PisEmptyProof (⇔)
          ↳15 TRUE
        ↳16 UsableRulesProof (⇔)
          ↳17 QDP
      ↳18 QDP
        ↳19 UsableRulesProof (⇔)
          ↳20 QDP
        ↳21 UsableRulesProof (⇔)
          ↳22 QDP
          ↳23 UsableRulesReductionPairsProof (⇔)
          ↳24 QDP
          ↳25 PisEmptyProof (⇔)
          ↳26 TRUE
      ↳27 QDP
        ↳28 UsableRulesProof (⇔)
          ↳29 QDP
          ↳30 UsableRulesReductionPairsProof (⇔)
          ↳31 QDP
          ↳32 PisEmptyProof (⇔)
          ↳33 TRUE
        ↳34 UsableRulesProof (⇔)
          ↳35 QDP
      ↳36 QDP
        ↳37 UsableRulesProof (⇔)
          ↳38 QDP
          ↳39 QDPOrderProof (⇔)
          ↳40 QDP
          ↳41 UsableRulesProof (⇔)
          ↳42 QDP
          ↳43 MRRProof (⇔)
          ↳44 QDP
          ↳45 PisEmptyProof (⇔)
          ↳46 TRUE
        ↳47 UsableRulesProof (⇔)
          ↳48 QDP

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

c'(active(x)) → c'(g(f(mark(x))))
g(f(active(x))) → g(mark(x))
c'(proper(x)) → c'(ok(x))
f(proper(x)) → proper(f(x))
g(proper(x)) → proper(g(x))
ok(f(x)) → f(ok(x))
ok(g(x)) → g(ok(x))
mark(top(x)) → proper(top(x))
ok(top(x)) → active(top(x))

Q is empty.

(3) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(4) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

c'(active(x)) → c'(g(f(mark(x))))
g(f(active(x))) → g(mark(x))
c'(proper(x)) → c'(ok(x))
f(proper(x)) → proper(f(x))
g(proper(x)) → proper(g(x))
ok(f(x)) → f(ok(x))
ok(g(x)) → g(ok(x))
mark(top(x)) → proper(top(x))
ok(top(x)) → active(top(x))

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVE(c) → F(g(c))
ACTIVE(c) → G(c)
PROPER(f(X)) → F(proper(X))
PROPER(f(X)) → PROPER(X)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
F(ok(X)) → F(X)
G(ok(X)) → G(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 6 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(ok(X)) → G(X)

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(10) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(ok(X)) → G(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

G(ok(X)) → G(X)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(G(x1)) = 2·x1   
POL(ok(x1)) = 2·x1   

(13) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(14) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(15) TRUE

(16) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(17) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(ok(X)) → G(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(ok(X)) → F(X)

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(19) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(ok(X)) → F(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(21) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(ok(X)) → F(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(23) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

F(ok(X)) → F(X)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(F(x1)) = 2·x1   
POL(ok(x1)) = 2·x1   

(24) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(26) TRUE

(27) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROPER(g(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(28) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(29) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROPER(g(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(30) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

PROPER(g(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(PROPER(x1)) = 2·x1   
POL(f(x1)) = 2·x1   
POL(g(x1)) = 2·x1   

(31) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(32) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(33) TRUE

(34) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(35) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROPER(g(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(36) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(37) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(38) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
g(ok(X)) → ok(g(X))
f(ok(X)) → ok(f(X))
active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TOP(mark(X)) → TOP(proper(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(TOP(x1)) =
/0A\
\0A/
 +
/0A1A\
\0A1A/
·x1

POL(ok(x1)) =
/-I\
\-I/
 +
/0A-I\
\-I0A/
·x1

POL(active(x1)) =
/0A\
\-I/
 +
/-I0A\
\-I0A/
·x1

POL(mark(x1)) =
/0A\
\0A/
 +
/0A0A\
\-I1A/
·x1

POL(proper(x1)) =
/0A\
\-I/
 +
/-I-I\
\-I0A/
·x1

POL(f(x1)) =
/0A\
\0A/
 +
/0A-I\
\0A0A/
·x1

POL(c) =
/0A\
\1A/

POL(g(x1)) =
/-I\
\-I/
 +
/0A-I\
\-I-I/
·x1

The following usable rules [FROCOS05] were oriented:

proper(f(X)) → f(proper(X))
proper(c) → ok(c)
active(c) → mark(f(g(c)))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
proper(g(X)) → g(proper(X))
active(f(g(X))) → mark(g(X))

(40) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
g(ok(X)) → ok(g(X))
f(ok(X)) → ok(f(X))
active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(41) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(42) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))
g(ok(X)) → ok(g(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(43) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

TOP(ok(X)) → TOP(active(X))

Strictly oriented rules of the TRS R:

active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))

Used ordering: Polynomial interpretation [POLO]:

POL(TOP(x1)) = x1   
POL(active(x1)) = 1 + x1   
POL(c) = 0   
POL(f(x1)) = x1   
POL(g(x1)) = x1   
POL(mark(x1)) = x1   
POL(ok(x1)) = 2 + x1   

(44) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

g(ok(X)) → ok(g(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(45) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(46) TRUE

(47) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(48) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

proper(c) → ok(c)
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
g(ok(X)) → ok(g(X))
f(ok(X)) → ok(f(X))
active(c) → mark(f(g(c)))
active(f(g(X))) → mark(g(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.